Find the Green function of $\Omega:=\left\{x\in\mathbb{R}^n:\lVert x\rVert<R, x_n>0\right\}$ and show that the function you've found is indeed a Green function! You are allowed to use the Green function of the whole ball $B_R(0):=\left\{x\in\mathbb{R}^n: \lVert x\rVert<R\right\}$ and of the upper half space $H:=\left\{x\in\mathbb{R}^n: x_n>0\right\}$ without proving their properties.
First of all, I wish you a happy new year.
Then I want to give you the Green functions of $B_R(0)$ and $H$ as suggested in the task.
1. Green function of $B_R(0)$: $$ G_n(x,y)=E_n(x-y)-E_n\left(\frac{\lVert y\rVert}{R}(x-y*)\right), $$ whereat $y*$ is defined by $$ y^*:=\frac{R^2}{\lVert y\rVert^2}y $$ for $y\in B_R(0)$ and $$ E_n(x):=\begin{cases}\frac{1}{2\pi}\ln\left(\frac{1}{\lVert x\rVert}\right), & n=2\\\frac{1}{(n-2)\sigma_n}\frac{1}{\lVert x\rVert^{n-2}}, & n>2\end{cases} $$ for all $x\in\mathbb{R}^n\setminus\left\{0\right\}$ is the fudamental solution of the Laplace equation.
2. Green function of $H$: $$ G_n(x,y):=E_n(x-y)-E_n(x-\overline{y}), $$ whereat for $y=(y_1,\ldots,y_n)\in H$ it is $\overline{y}:=(y_1,\ldots,y_{n-1},-y_n)$.
Both Green function were derived from reflection arguments. The first one by refelection at the sphere, the second by reflection at the plain.
3. Green function of $\Omega$:
The idea I used is again reflection. Reflecting a point $y$ in $\Omega$ at the sphere (getting $y^*$ that is outside of $\Omega$, but lying in $H$), reflecting $y$ at the plane (getting $\overline{y}$, lying in the lower half ball) and reflecting $\overline{y}$ at the sphere of the lower half ball (getting $\overline{y}^*$, lying in the lower half space, outside of the lower half ball).
Then for the Green function of $\Omega$, using the Green functions above, I get $$ G_n(x,y):=E_n(x-y)-E_n\left(\frac{\lVert y\rVert}{R}(x-y^*)\right)-E_n(x-\overline{y})+E_n\left(\frac{\lVert\overline{y}\rVert}{R}(x-\overline{y}^*)\right). $$ Define $$ g_n(x,y):=E_n\left(\frac{\lVert y\rVert}{R}(x-y^*)\right)+E_n(x-\overline{y})-E_n\left(\frac{\lVert\overline{y}\rVert}{R}(x-\overline{y}^*)\right), $$ then it is $$ G_n(x,y)=E_n(x-y)-g_n(x,y). $$
Now there are two things to show.
(1) $\forall y\in\Omega~\forall x\in\partial\Omega: G_n(x,y)=0$
(2) $\forall y\in\Omega: g_n(\cdot,y)\in C(\overline{\Omega})\cap C^2(\Omega)$ and $g_n(\cdot,y)$ is harmonic in $\Omega$
ad (1)
Because the bound consists of two parts, namely $T:=S_R(0)\cap H$ and $W:=\left\{x\in\mathbb{R}^n: x_n=0\right\}\cap B_R(0)$, two cases have two be considered:
I. Consider any $y\in\Omega$ and any $x\in T$.
Then using the properties of the Green function of $B_R(0)$, it is directly $$ E_n(x-y)-E_n\left(\frac{\lVert y\rVert}{R}(x-y^*)\right)=0,~~~~~ E_n(x-\overline{y})=E_n\left(\frac{\lVert\overline{y}\rVert}{R}(x-\overline{y}^*)\right), $$ i.e. $G_n(x,y)=0$.
II. Consider any $y\in\Omega$ and any $x\in W$.
Then using the properties of the Green function of $H$, it is directly $$ E_n(x-y)-E_n(x-\overline{y})=0 $$ and moreover it is $$ -E_n\left(\frac{\lVert y\rVert}{R}(x-y^*)\right)+E_n\left(\frac{\lVert\overline{y}\rVert}{R}(x-\overline{y}^*)\right)\\=-E_n\left(\frac{\lVert y\rVert}{R}x-\frac{R}{\lVert y\rVert}y\right)+E_n\left(\frac{\lVert\overline{y}\rVert}{R}x-\frac{R}{\lVert\overline{y}\rVert}\overline{y}\right). $$ Because $\lVert y\rVert=\lVert\overline{y}\rVert$, $\frac{\lVert y\rVert}{R}x\in\left\{x\in\mathbb{R}^n: x_n=0\right\}$ and $\frac{R}{\lVert y\rVert}y\in H$, this is $0$, because of the properties of the Green function of $H$.
So it is $G_n(x,y)=0$.
ad (2)
Because $E_n$ is harmonic in $\mathbb{R}^n\setminus\left\{0\right\}$, it is especially harmonic in $\Omega$. So for any $y\in\Omega$, $g_n(\cdot,y)$ is harmonic in $\Omega$, because it is only a sum of different harmonic functions on $\Omega$ then.
For the same reason for any $y\in\Omega$ $g_n(\cdot,y)$ is continious on $\overline{\Omega}$ (because $E_n$ is continious on $\overline{\Omega}$ and $g_n(\cdot,y)$ is a sum of functions that are continious on $\overline{\Omega}$ then).
And again: $E_n$ is twice continiously differentiable on $\Omega$. So $g_n(\cdot,y)$ - as a sum of functions, that are twice continiously differentiable on $\Omega$ - is twice continiously differentiable on $\Omega$.
That's my answer. Could you please say me if it is allright?
Do I have to add more details or is it enough?
With greetings!
math12
